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InterviewSolution
This section includes InterviewSolutions, each offering curated multiple-choice questions to sharpen your knowledge and support exam preparation. Choose a topic below to get started.
| 751. |
Which type of bond is not present in `HNO_2` molecule?A. CovalentB. Co-ordinateC. lonicD. lonic as well as co-ordinate. |
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Answer» Correct Answer - D H-O-N=O does not contain on and co ordinate bond. |
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| 752. |
The molecule having smallest bond angle isA. `NCl_(3)`B. `AsCl_3`C. `SbCl_(3)`D. `PCl_(3)` |
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Answer» Correct Answer - C As electronegativity of central atom decreases, bond angle decreases. (Hybridisation and number of lone pair on cental atom are same in all options) |
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| 753. |
Which one in the following contains ionic as well as covalent bondA. `CH_(4)`B. `H_2`C. KCND. KCl |
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Answer» Correct Answer - C Structure of KCN is `[K^(+)(C^(-)-=ddot(N))]` |
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| 754. |
Which one of the following has the smallest bond angle?A. `NH_3`B. `BeF_2`C. `H_2O`D. `CH_4` |
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Answer» Correct Answer - C Hybridisation involved in `NH_3` is `sp^3` with one lone pair, in `BeF_2` is `sp^2` with no lone pair, in `H_2O` is `sp^3` with two lone pairs, in `CH_4` is `sp^3` with no lone pair, in `SO_2` is `sp^2` with one lone pair. Therefore, least bond angle is `H_2O` `(~~104-5^(@))` |
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| 755. |
Which one of the following molecules has the smallest bond angle ?A. `NH_(3)`B. `PH_(3)`C. `H_(2)O`D. `H_(2)Se` |
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Answer» Correct Answer - D `NH_3 =107^(@), PH_(3)=93^(@),H_(2)O=104.5^(@)` `H_(2)Se=91^(@),H_(2)S=92.5^(@)` |
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| 756. |
Which of the following compounds has the smallest bond angle?A. `H_(2)O`B. `H_(2)S`C. `NH_(3)`D. `CO_(2)` |
| Answer» (b) Bond angle of `H_(2)S` is smallest because S-atom is large in size and has low electronegativity. | |
| 757. |
For the reaction `M^(x+)+MnO_(4)^(ө)rarrMO_(3)^(ө)+Mn^(2+)+(1//2)O_(2)` if `1 "mol of" MnO_(4)^(ө)` oxidises `1.67 "mol of" M^(x+) "to" MO_(3)^(ө)`, then the value of `x` in the reaction isA. 5B. 3C. 2D. 1 |
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Answer» Correct Answer - C 1 mole of `Mn^(+7)O_4^(-)` changes of `Mn^(+2)` by gaining of 5 electrons. i.e., 5 moles of electrons are lost by 1.67 moles of `M^(x+)` `therefore` 1 mole of `M^(x+)` will lose electrons `=5/1.67=3.01` Since `M^(x+)` changes to `MO_3^(-)` where oxidation state of M= +5 . |
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| 758. |
The shapes of `PCl_4^+, PCl_4^-` and `AsCl_4` are respectively :A. square planar , tetrahedral , see -sawB. tetrahedral , see-saw , trigonal bipyramidalC. tetrahedral, square planar and pentagonal bipyramidalD. trigonal bipyramidal tetrahedral and square pyramidal |
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Answer» Correct Answer - B `PCl_4^+ (sp^3) , PCl_4^(-) (sp^3 d," one lone pair")` , ` AsCl_5 (sp^3 d," no lone pair")` |
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| 759. |
`PCl_5` in solid state exists as `PCl_4^+` and `PCl_6^-` Also in some solvents it undergoes dissoctation as `2PCl_5 hArr PCl_2^+ +PCl_6^-`. Which statement is wrong ?A. In `PCl_5`, all the `P -Cl` bonds are of same energyB. `PCl_5` has no lone pair of electronC. `PCl_5` is a white solid which melts at `167^(@)C`D. ` PCl_5` gives white fumess with moist air |
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Answer» Correct Answer - A ` PC l_S` has ` 3 sigma` bond at equatorial and two at axial position relatively weaker . |
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| 760. |
Statement : The bond energy of `P-Cl` bond in `PCl_3` and `PCl_5` different . Explanation : In `PCl_3 sp^3 -p` overlapping whereas in `PCl_5 sp^3 d-p` overlapping in noticed .A. `S` is correct by ` E` is wrong .B. `S` is wrong but `E` is wrong .C. Both `S` and `E` are correct and `E` is correct explanation os ` S` .D. Both `S` and ` E` are conrrect and `E` is correct explanation os `S`. |
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Answer» Correct Answer - C Explanation is correct reason for statement . |
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| 761. |
` PCl_5` exists but ` NCl_5` does not because :A. nitrogen has no vacant ` 2d-`orbitalsB. `NCl_5` is unstableC. nitrogen atoms is much smaller than phosphorusD. nitrogen is highly inert |
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Answer» Correct Answer - A Excitation of 2s-electron in N is not possible . |
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| 762. |
A hybrid orbital formed from s and p-orbital can contribute toA. a `sigma` bond onlyB. `pi`-bond onlyC. either `sigma` or `pi` bondD. cannot be predicted |
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Answer» Correct Answer - A s-p bond involves axial overlapping. Therefore, it is a sigma `(sigma)` bond. |
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| 763. |
Paramagnetism is shown by the molecules which haveA. paired electronsB. unpaired electronsC. lone pair of electronsD. bond order more than one. |
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Answer» Correct Answer - B Presence of unpaired electrons makes a molecule paramagnetic . More the number of unpaired electrons, higher is the paramagnetism. |
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| 764. |
Which of the following formulae does not show the correct relationship ?A. `B.O.=1/2(N_(b)-N_(a))`B. `B.O.propto1/("Bond length")`C. `B.O.propto1/"Bond dissociation energy"`D. `N_(b) gt N_(a), B.O. = + ve` |
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Answer» Correct Answer - C Bond order ` propto` Bond dissociation energy. |
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| 765. |
Comprehension given below is followed by some multiple choice question, Each question has one correct options. Choose the correct option. Molecular orbitals are formed by the overlap of atomic orbitals. Two atomic orbitals combine to form two molecular orbitals called bonding molecular orbital (BMO) and anti-bonding molecular orbital (ABMO). Energy of anti-bonding orbital is raised above the parent atomic orbitals that have combined and hte energy of the bonding orbital is lowered than the parent atomic orbitals. energies of various molecular orbitals for elements hydrogen to nitrogen increase in the order `sigma1s lt sigma^(star)1s lt sigma^(star)2s lt ((pi2p_(x))=(pi2p_(y))) lt sigma2p_(z) lt (pi^(star)2p_(x) = pi^(star)2p_(y)) lt sigma^(star)2p_(z)` and For oxygen and fluorine order of enregy of molecules orbitals is given below. `sigma1s lt sigma^(star)1s lt sigma2s lt sigma^(star)2s lt sigmap_(z) lt (pi2p_(x) ~~ pi2p_(y)) lt (pi^(star)2p_(x)~~ pi^(star)2py) lt sigma^(star)2p_(z)` Different atomic orbitalsof one atom combine with those atoms orbitals of the second atom which have comparable energies and proper orientation. Further, if the overlapping is head on, the molecular orbital is called sigma, `sigma` andif the overlap is lateral, the molecular orbital is called pi, `pi`. The molecular orbitals are filled with electrons according to the same rules as followed for filling of atomic orbitals. However, the order for filling is not the same for all molecules or their ions. Bond order is one of the most important parameters to compare the strength of bonds. 67) Which of the following pair is expected to have the same bonod order?A. `C_(2)`B. `B_(2)`C. `O_(2)`D. `Be_(2)` |
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Answer» Correct Answer - C For all elements which have atomic number more than 7 (beyond nitrogen ) the energy of `sigma2p_(z)` is lower than `pi 2p_(x) and pi 2p_(y)`- orbitals. |
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| 766. |
Which of the following relationships is true?A. Bond dissociation energy of `O_(2) and O_(2)^(-)` are same.B. Bond dissociation energy of `O_(2)^(+)` is higher than `O_(2)`,C. Bond dissociation energy of `O_(2)^(-) and O_(2)^(2-)` are same.D. Bond dissociation energy of `O_(2)^(2-) ` is higher than `O_(2)^(-)`. |
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Answer» Correct Answer - B M.O. configuration of `O_(2)^(+)`: `(sigma 1s^(2))(sigma^(**)1s^(2))(sigma2s^(2))(sigma^(**)2s^(2))(sigma2p_(z)^(2))(pi 2p_(x)^(2) = pi 2p_(y)^(2))(pi^(**)2p_(x)^(1))` B.O. ` = (10-5)/2 = 2.5` B.O. of `O_(2) = 2 {:[(O_(2)^(+),gt,O_(2),gt,O_(2)^(-),gt,O_(2)^(2-)),(2.5,,2,,1.5,,1)]:}` |
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| 767. |
The interionic attraction depends on interaction ofA. solute-soluteB. solvent-solventC. the chargesD. molecular properties |
| Answer» Correct Answer - C | |
| 768. |
The bonds present in `N_(2)O_(4)` areA. only ionicB. only covalentC. covalent and coordinateD. ionic and coordinate |
| Answer» Correct Answer - C | |
| 769. |
Fill in the blanks with appropriate choice. Bond ordr of `N_(2)^(+) is ul(" "P)`while that of `N_(2) is ul(" "Q)`. Bond order of `O_(2)^(+) is ul(" "R)` while that of `O_(2) is ul(" "S)`. N - N bond distance `ul(" "T)` when `N_(2)` changes to `N_(2)^(+) ` and when `O_(2)` changes to `O_(2)^(+)` , the O - O bond distance `ul(" "U)`.A. `{:(P,Q,R,S," "T," "U),(2,2.5,2.5,1,"increases","decreases"):}`B. `{:(P,Q,R,S," "T," "U),(2.5,3,2,1.5,"decreases","increases"):}`C. `{:(P,Q,R,S," "T," "U),(3,2,1.5,1,"increases","decreases"):}`D. `{:(P,Q,R,S," "T," "U),(2.5,3,2.5,2,"increases","decreases"):}` |
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Answer» Correct Answer - D `N_(2)^(+) : (sigma1s^(2))(sigma^(**)1s^(2))(sigma2s^(2))(sigma^(**)2s^(2))(pi 2p_(x)^(2) = pi 2p_(y)^(2))(sigma2p_(z)^(1))` B.O. ` = 1/2 xx (9-4) = 2.5` `N_(2) : B.O. = 1/2 xx (10 - 4) = 3` `O_(2)^(+) : (sigma1s^(2))(sigma^(**)1s^(2))(sigma2s^(2)) (sigma^(**)2s^(2))(sigma2p_(z)^(2))(pi 2p_(x)^(2) pi 2p_(y)^(2))(pi^(**)2p_(x)^(1))` B.O. ` = 1/2 xx (10 - 5) = 2.5` ` O_(2) : B.O. = 1/2 xx (10 - 6) = 2` Since `N_(2)^(+)` has lower bond order than `N_(2)`, bond length of N - N in `N_(2)^(+)` increases. In `O_(2)^(+)`, bond order increases from 2 to `2.5` hence, bond length decreases. |
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| 770. |
When `N_(2)` goes to `N_(2)^(+)`, the `N-N` bond distance …………, and when `O_(2)` goes to `O_(2)^(+)` the `O-O` bond distance…….A. Decreases , increasesB. Increases, decreasesC. Increases, increasesD. None of these |
| Answer» Correct Answer - B | |
| 771. |
Polarization of electrons in acrolein may be written as:A. `overset(delta^(-))(C)H_(2)=CH-overset(detal^(+))(C)H=O`B. `overset(delta^(-))(C)H_(2)=CH-CH=overset(delta^(+))(O)`C. `overset(delta^(-))(C)H_(2)=overset(delta^(+))(C)H-CH=O`D. `overset(delta^(+))(C)H_(2)=CH-CH=overset(delta^(-))(O)` |
| Answer» Correct Answer - D | |
| 772. |
Why do `NH_(3)` and `H_(2)O` act as electron pair donors? |
| Answer» Ammonia molecule possesses one lone pair on nitrogen atom. Water molecule possesses two lone pairs on oxygen atom. They can therefore donate the excess electron pair to any ion or atom or molecule that has electron defficiency. | |
| 773. |
Between `NF_(3)` and `BF_(3)`, which is the more polar molecule? Why? |
| Answer» Between `NF_(3)` and `BF_(3), NF_(3)` is polar but `BF_(3)` is non-polar. In the `BF_(3)` molecule there is no lone pair of electrons on the central atom. The shape of this molecule is trigonal planar that is symmetrical. `NF_(3)` has three bond pairs and one lone pair. But due to the presence of a lone pair on nitrogen the polarities are not cancelled. Hence, the `NF_(3)` molecule is polar. | |
| 774. |
Among `N_(2)O,SO_(2),I_(3)^(+)` and `I_(3)^(-)`, the linear species are ………..and……….A. `NO_(2)^(-),I_(3)^(+),H_(2)O`B. `N_(2)O,I_(3)^(+),N_(3)^(-)`C. `N_(2)O,I_(3)^(-),N_(3)^(-)`D. `N^(3-),I^(3+),SO_(2)` |
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Answer» Correct Answer - C `N_(2)O: bar(N)=overset(+)(N) =O` `I_(3)^(-)`: With 2 bp and 3 lp around central atom is linear `N_(3)^(-)`: with 2 bp (stereoactive pairs) around the central atom is linear. |
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| 775. |
Among `N_(2)O,SO_(2),I_(3)^(+)` and `I_(3)^(-)`, the linear species are ………..and………. |
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Answer» Correct Answer - `N_(2)O,I_(3)^(-)` |
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| 776. |
The hybridisation present in `IF_(3)` isA. `sp^(3)d`B. `sp^(3)`C. `sp^(3)d^(2)`D. `sp^(3)d^(3)` |
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Answer» Correct Answer - A Hybridisation present in `IF_(3)` is `sp^(3)d`. In `IF_(3)`, 3 bond pairs of electrons and 2 lone pair of electrons are present. |
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| 777. |
Match each of the diatomic molecules in Column I with its property properties in Column II A. `{:(A,B,C,D),(q","r","s,q","r","s","t,q","r","t,p","q","t):}`B. `{:(A,B,C,D),(p","q","r","t,q","r","s","t,p","q","r","t,p","r","s","t):}`C. `{:(A,B,C,D),(q","r","s","t,p","q","r,r","s","t,p","q","r","t):}`D. `{:(A,B,C,D),(q","r","s,q","r","s","t,q","r","t,p","q","t):}` |
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Answer» Correct Answer - `Atop,q,r,t;Btoq,r,s,t;Ctop,q,r,t;Dtop,r,s,t |
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| 778. |
Match the following columns and choose the correct option from the codes given below. A. `{:(,,A,B,C,),(,,1,1,2,):}`B. `{:(,,A,B,C,),(,,2,2,1,):}`C. `{:(,,A,B,C,),(,,1,2,1,):}`D. `{:(,,A,B,C,),(,,2,1,2,):}` |
| Answer» Correct Answer - (c ) | |
| 779. |
Which of the following is a polar compound ?A. HClB. `H_(2)Se`C. `CH_(4)`D. HI |
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Answer» Correct Answer - A Higher is the difference in electronegativity of two covalently bonded atoms, higher is the polarity . In HCl there is high difference in the electronegativity of H and Cl atom so it is a polar compound. |
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| 780. |
Which of the following is the most polar?A. `C Cl_(4)`B. `CHCl_(3)`C. `CH_(3)OH`D. `CH_(3)Cl` |
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Answer» Correct Answer - C Dipole moment of `CH_(3)OH` is maximum in it. |
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| 781. |
Which of the following has zero dipole moment ?A. `CH_(2)Cl_(2)`B. `CH_(4)`C. `NH_(3)`D. `PH_(3)` |
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Answer» Correct Answer - B `CH_(4)` have regular tetrahedron so its dipole moment is zero. |
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| 782. |
The dipole moment of chlorobenzene is 1.73 D . The dipole moment of p-dichlorobenzene is expected to beA. 3.46 DB. 0.00DC. 1.73 DD. 1.00 D |
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Answer» Correct Answer - B Due to symmetry dipole moment of p-dichloro benzene is zero. |
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| 783. |
Which is the most covalent?A. C-FB. C-OC. C-SD. C-Br |
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Answer» (c ) ` because "Convalent character" prop (1)/("Ionic character") prop (1)/("difference in electronegativity")` `:.` C-S is the most covalent. |
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| 784. |
Other factors being constant which bond order is expected to correspond to shortest bond lengthA. `2 1/2`B. `1 1/2`C. 2D. 0.5 |
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Answer» Correct Answer - A More is bond order, shorter will be the bond length `("B.O."prop 1/(B.L.))` |
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| 785. |
A polar covalent bond with positive and negative charge centres at its ends is called a dipole. The polarity of a dipole is measured by its dipole moment. Mathematically it is expressed as dipole moment, `mu=q xx d` where q and d are the net charge and the distance between the two charges respectively. Dipole moment is a vector quantity. The net dipole moment of a polyatomic molecule is the resultant of the various bond moments present in the molecule. The values of dipole moment are expressed in Debye (D) or in SI units in terms of coulomb- metre (Cm). One of the most important applications of dipole moment is in the determination of geometry and shape of molecules besides prediction of a number of properties of the molecules. The dipole moment of HBr is `0.78 xx 10^(-18)` esu cm. The bond length of HBr is 1.41 Ã…. The percentage ionic character of HBr isA. 7.54B. 11.52C. 15.7D. 27.3 |
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Answer» Correct Answer - B % ionic character `=(0.78xx10^(-18)xx100)/(1.41xx10^(-8)xx4.8xx10^(-10))=11.52%` |
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| 786. |
A polar covalent bond with positive and negative charge centres at its ends is called a dipole. The polarity of a dipole is measured by its dipole moment. Mathematically it is expressed as dipole moment, `mu=q xx d` where q and d are the net charge and the distance between the two charges respectively. Dipole moment is a vector quantity. The net dipole moment of a polyatomic molecule is the resultant of the various bond moments present in the molecule. The values of dipole moment are expressed in Debye (D) or in SI units in terms of coulomb- metre (Cm). One of the most important applications of dipole moment is in the determination of geometry and shape of molecules besides prediction of a number of properties of the molecules. A diatomic molecule has a dipole of 1.2 D, if the bond distance is 1 Ã…, what percentage of electronic charge exists on each atom.A. 25% of eB. 29% of eC. 19% of eD. 20% of e. |
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Answer» Correct Answer - A `mu=e/d=(1.2D)/(1.0A)=(1.2xx10^(-18))/(1xx10^(-18)"cm")=1.2xx10^(-10)` esu `therefore` % of e=`(1.2xx10^(-10))/(4.8xx10^(-10))xx100`=25% |
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| 787. |
The species `CO, CN^(-)`,and `NO^(+)` which are isoelectronic with `N_(2)`, are much more reactive than `N_(2)` becauseA. COB. `N_(2)^(+)`C. `N_(2)^(-)`D. `O_(2)^(+)` |
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Answer» Correct Answer - A Number of electrons in `CN^(-)=14` Number of electron in CO=14 Number of electron in `N_(2)^(+)=13` Number of electrons in `N_(2)^(-)=15` Number of electron in `O_(2)^(+)` =15 |
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| 788. |
Which of the following statements regarding covalent bond is not trueA. The electrons are shared between atomsB. the bond is non-directionalC. the strength of the bond depends upon the extent of overlapingD. the bond formed may or may not be polar |
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Answer» Correct Answer - B Covalent bond is directional. |
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| 789. |
The electronic configuration of metal M is `1s^(2)2s^(2)2p^(6)3s^(1).` The formula of its oxide will be :A. MOB. `M_(2)O`C. `M_(2)O_(3)`D. `MO_(2)` |
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Answer» Correct Answer - B The electronic configuration of Na(Z=11) is `1s^(2),2s^(2)2p^(6),3s^(1)` . The oxide of Na is `Na_(2)O` |
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| 790. |
The electronic configuration of four elements `L, P, Q` and `R` are given in brackets `L (1s^(2), 2s^(2), 2p^(4)) , P (1s^(2), 2p^(6), 3s^(1))` `Q (1s^(2), 2s^(2)2p^(6), 3s^(2)3p^(5)) , R (1s^(2), 2s^(2)2p^(6), 3s^(2))` The formula of ionic compounds that can be formed between elements areA. `L_2P,RL` ,PQ and `R_(2)Q`B. `LP,RL,PQ` and RQC. `P_(2)L`,RL,PQ and `RQ_(2)`D. `LP,R_(2)L,P_(2)Q` and RQ |
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Answer» Correct Answer - C valencies of L, Q,P and R is -2,-1,+1 and +2 respectively and one electron in their outermost shell so they will form cation easily. |
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| 791. |
The electronic configuration of four atoms are given in brackets : `L(1s^(2)2s^(2)2p^(1))," "M(1s^(2)2s^(2)2p^(5)),` `Q(1s^(2)2s^(2)2p^(6)3s^(1))," "R(1s^(2)2s^(2)2p^(2)),` The element that would most readily form a diatomic molecule isA. QB. MC. RD. L |
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Answer» Correct Answer - B By sharing of 1 electron each, both atoms of M will get a completed octet. |
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| 792. |
The type of hybrid orbitals used by the chlorine atom in `CIO_(2^(-))` isA. `sp^(3)`B. `sp^(2)`C. `sp`D. None of these |
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Answer» Correct Answer - a |
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| 793. |
The Type of hybrid orbitals used by the chlorine atom in `CIO_(2)^(Theta)` is .A. `sp^3`B. `sp^2`C. `sp`D. None |
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Answer» Correct Answer - A In `ClO_2^(-)` ion H=`1/2`[7+0-0+1]=`1/2xx8`=4 so hybridisation is `sp^3` |
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| 794. |
Which of the following chloride has considerable covalent character ?A. LiClB. NaClC. KClD. CsCl. |
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Answer» Correct Answer - A `Li^+` being smaller has highest polarising power according to Fazan rules. Hence LiCl should have maximum ionic character. |
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| 795. |
The hybrid state of C in `CS_2` should beA. `sp^2`B. spC. `sp^3`D. no specific |
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Answer» Correct Answer - B S=C=S (linear) is best explained by sp-hybrid state of C. |
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| 796. |
The molecule which has pyramidal shape isA. `PCl_(3)`B. `SO_(3)`C. `CO_(3)^(2-)`D. `NO_(3)^(-)` |
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Answer» Correct Answer - A `sp^(3)` hybridization of P atom with one orbital occupied by one lone pair of electrons. |
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| 797. |
The molecule which has zero dipole moment isA. `CH_(2)Cl_(2)`B. `BF_(3)`C. `NF_(3)`D. `ClO_(2)` |
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Answer» Correct Answer - A As `CH_(2)Cl_(2)` has, the dipole moment zero because of the two Cl atoms cancel out. |
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| 798. |
The melecule that has linear structure is:A. `CO_(2)`B. `NO_(2)`C. `SO_(2)`D. `SiO_(2)` |
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Answer» Correct Answer - a |
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| 799. |
Which of the following will have zero dipole moment?A. 1,1-dichloroethyleneB. cis -1,2-dichloroethyleneC. trans-1,2,-dichloroethyleneD. none of the above |
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Answer» Correct Answer - c |
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| 800. |
The element that has the strongest metallic bond among `._(11)A^(23), ._(12)B^(24), ._(13)C^(27)` and `._(19)D^(39)` is ______.A. AB. BC. CD. D |
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Answer» `{:(,"Atomic","Electronic"),("Elements","numbers","Configurations"),("A",11,"2, 8, 1"),("B",12,"2, 8, 2"),("C",13,"2, 8, 3"),("D",19,"2, 8, 8, 1"):}` A and D can lose one electrons and become stable, B loses two electrons and C loses 3. Strength of metallic bond depends on the charges on metal kernels and radii. As C has the maximum charge and is the smallest in size among the given elements, the metallic bond is the strongest in C. |
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